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NCERT Solutions for Exercise 6.2 Class 10 Maths Chapter 6 - Triangles

NCERT Solutions for Exercise 6.2 Class 10 Maths Chapter 6 - Triangles

Updated on Apr 29, 2025 05:39 PM IST | #CBSE Class 10th

Similarity of figures is a property that is used to compare figures. Two triangles are said to be similar if their corresponding sides are in equal ratio and if their corresponding angles are equal. If the corresponding sides of two triangles are equal, then these triangles are called equiangular triangles. In an equiangular triangle, the ratio of the corresponding sides of the two sides is always the same. This property of triangles follows from the basic proportionality theorem.

This Story also Contains
  1. Download Free PDF of NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.2
  2. Assess NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.2
  3. Topics Covered in Chapter 6, Triangles: Exercise 6.2
  4. NCERT Solutions Subject Wise
  5. Subject-Wise NCERT Exemplar Solutions
NCERT Solutions for Exercise 6.2 Class 10 Maths Chapter 6 - Triangles
NCERT Solutions for Exercise 6.2 Class 10 Maths Chapter 6 - Triangles
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These NCERT solutions of Class 10 maths ex 6.2 contain two theorems: (Theorem 1) When a parallel line is drawn along one of the triangle's sides and intersects the other two sides at certain points, the two sides are divided in the same proportion. (Theorem 2) Any line that divides the two sides of a triangle in the same ratio is parallel to the third side. NCERT solutions for exercise 6.2 Class 10 Maths chapter 6 Triangles discuss the concept as per the NCERT Book, like a triangle as a polygon, the interior angles of a triangle, the exterior angles of a triangle, Pythagoras' theorem, and conditions of similarity in triangles.

Download Free PDF of NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.2

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Assess NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.2

Triangles Class 10 Chapter 6 Exercise: 6.2

Q1 In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

1635923248436

Answer:

(i)

Let EC be x

Given: DE || BC

By using the proportionality theorem, we get

ADDB=AEEC

1.53=1x

x=31.5=2cm

EC=2cm

(ii)

Let AD be x

Given: DE || BC

By using the proportionality theorem, we get

ADDB=AEEC

x7.2=1.85.4

x=7.23=2.4cm

AD=2.4cm

Q2 (1) E and F are points on the sides PQ and PR respectively of a PQR. For each of the following cases, state whether EF || QR: PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

Answer:

(i)a

Given :

PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

PEEQ=3.93=1.3cm and PFFR=3.62.4=1.5cm

We have

PEEQPFFR

Hence, EF is not parallel to QR.

Q2 (2) E and F are points on the sides PQ and PR respectively of a triangle PQR. For each of the following cases, state whether EF || QR : PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm

Answer:

(ii)b

Given :

PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm

PEEQ=44.5=89cm and PFFR=89cm

We have

PEEQ=PFFR

Hence, EF is parallel to QR.

Q2 (3) E and F are points on the sides PQ and PR respectively of a triangle PQR. For each of the following cases, state whether EF || QR : PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

Answer:

(iii)

c
Given:

PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

PEPQ=0.181.28=964cm and PFPR=0.362.56=964cm

We have

PEEQ=PFFR

Hence, EF is parallel to QR.

Q3 In Fig. 6.18, if LM || CB and LN || CD, prove that AMAB=ANAD

1635923547862

Answer:

Given : LM || CB and LN || CD

To prove :

AMAB=ANAD

Since , LM || CB so we have

AMAB=ALAC............(1)

Also, LN || CD

ALAC=ANAD..............(2)

From equations 1 and 2, we have

AMAB=ANAD

Hence proved.

Q4 In Fig. 6.19, DE || AC and DF || AE. Prove that BF / FE = BE / EC

1635923629376

Answer:

Given: DE || AC and DF || AE.

To prove:

BFFE=BEEC

Since , DE || AC so we have

BDDA=BEEC...........(1)

Also, DF || AE

BDDA=BFFE.............(2)

From equations (1) and (2), we have

BFFE=BEEC

Hence proved.

Q5 In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR.

1635923658934

Answer:

Given: DE || OQ and DF || OR.

To prove EF || QR.

Since DE || OQ, we have

PEEQ=PDDO..........(1)

Also, DF || OR

PFFR=PDDO.............(2)

From equations (1) and (2), we have

PEEQ=PFFR

Thus, EF || QR. (Converse of the basic proportionality theorem)

Hence proved.

Q6 In Fig. 6.21, A, B, and C are points on OP, OQ and OR, respectively, such that AB || PQ and AC || PR. Show that BC || QR.

1635923722792

Answer:

Given: AB || PQ and AC || PR

To prove: BC || QR

Since, AB || PQ so we have

OAAP=OBBQ............(1)

Also, AC || PR

OAAP=OCCR..............(2)

From equations (1) and (2), we have

OBBQ=OCCR

Therefore, BC || QR. (converse basic proportionality theorem)

Hence proved.

Q7 Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).

Answer:

d

Let PQ be a line passing through the midpoint of line AB and parallel to line BC, intersecting line AC at point Q.

i.e. PQ||BC and AP=PB .

Using the basic proportionality theorem, we have

APPB=AQQC.........(1)

Since AP=PB

AQQC=11

AQ=QC

Q is the midpoint of AC.

Q8 Using Theorem 6.2, prove that the line joining the midpoints of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).

Answer:

1745673977453

Let P is the midpoint of line AB, and Q is the midpoint of line AC.

PQ is the line joining midpoints P and Q of lines AB and AC, respectively.

i.e. AQ=QC and AP=PB .

We have,

APPB=11..........................1

AQQC=11...................................2

From equations (1) and (2), we get

AQQC=APPB

Therefore, by the basic proportionality theorem, we have PQ||BC

Q9 ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show
that AOBO=CODO

Answer:
e

Draw a line EF passing through point O such that EO||CDandFO||CD

To prove:

AOBO=CODO

In ADC , we have CD||EO

So, by using the basic proportionality theorem,

AEED=AOOC........................................1

In ABD , we have AB||EO

So, by using the basic proportionality theorem,

DEEA=ODBO........................................2

Using equations (1) and (2), we get

AOOC=BOOD

AOBO=CODO

Hence proved.

Q10 The diagonals of a quadrilateral ABCD intersect each other at point O such that AOBO=CODO Show that ABCD is a trapezium.

Answer:e

Draw a line EF passing through point O such that EO||AB

Given :

AOBO=CODO

In ABD , we have AB||EO

So, by using the basic proportionality theorem,

AEED=BODO............(1)

However, it is given that

AOCO=BODO.............(2)

Using equations (1) and (2), we get

AEED=AOCO

EO||CD (By basic proportionality theorem)

AB||EO||CD

AB||CD

Therefore, ABCD is a trapezium.

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JEE Main high scoring chapters and topics

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Topics Covered in Chapter 6, Triangles: Exercise 6.2

1. Similar Figures: Two figures are said to be similar if their corresponding sides are in the same proportion and corresponding angles are equal.

2. Conditions for Similar Triangles: The criteria for similarity of the triangle are AA (Angle-Angle), in which two angles of one triangle are equal to the corresponding two angles of another triangle.

3. Midpoint Theorem: In this theorem, the line segment that joins the midpoint of the two sides of the triangle is parallel to the third side of the triangle, and the length of this line segment is half of the third side of the triangle.

4. Parallel Lines and Transversals: A transversal divides the sides of the triangle into proportional segments when it intersects the two parallel lines.

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NCERT Solutions Subject Wise

Students must check the NCERT solutions for Class 10 Maths and Science given below:

JEE Main Important Mathematics Formulas

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

Subject-Wise NCERT Exemplar Solutions

Students must check the NCERT exemplar solutions for Class 10 Maths and Science given below:

Frequently Asked Questions (FAQs)

1. What are the figures that are similar?

The term "similar" refers to two figures that have the same shape but differ in size.

2. Difference between congruent and similar figures?

The congruent figures are all similar, but the opposite is not true.

3. What is the use of side ratios for similar triangle?

If we know the side ratio and the length of any side, we can find the length of the corresponding side of the other triangle by using the ratio.

4. What is the ratio of the area of triangles if the ratio of the sides is given?

The ratio of the areas of the triangles is equal to the square of the ratio of the sides.

5. If the ratio of the sides of two similar triangle is 1:2 then the ratio of their areas is?

The ratio of the areas of the triangles is equal to the square of the ratio of the sides.

Ratio of area = 1:4

6. If the ratio of the sides is given as 1:3 and the shorter side has length of 4cm then the length of the longer side is?

4/x=1/3

x=12cm

The length of the longer side is 12cm.

7. Mention one of the properties of the line that passes through the mid-point of any two sides of the triangle?

It will always be parallel to the third side of the triangle. 

8. Is similarity a sign of agreement between the figures?

No, resemblance only relates to the shape of the figures and does not imply equality.

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Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

Hello

Since you are a domicile of Karnataka and have studied under the Karnataka State Board for 11th and 12th , you are eligible for Karnataka State Quota for admission to various colleges in the state.

1. KCET (Karnataka Common Entrance Test): You must appear for the KCET exam, which is required for admission to undergraduate professional courses like engineering, medical, and other streams. Your exam score and rank will determine your eligibility for counseling.

2. Minority Income under 5 Lakh : If you are from a minority community and your family's income is below 5 lakh, you may be eligible for fee concessions or other benefits depending on the specific institution. Some colleges offer reservations or other advantages for students in this category.

3. Counseling and Seat Allocation:

After the KCET exam, you will need to participate in online counseling.

You need to select your preferred colleges and courses.

Seat allocation will be based on your rank , the availability of seats in your chosen colleges and your preferences.

4. Required Documents :

Domicile Certificate (proof that you are a resident of Karnataka).

Income Certificate (for minority category benefits).

Marksheets (11th and 12th from the Karnataka State Board).

KCET Admit Card and Scorecard.

This process will allow you to secure a seat based on your KCET performance and your category .

check link for more details

https://medicine.careers360.com/neet-college-predictor

Hope this helps you .

Hello Aspirant,  Hope your doing great,  your question was incomplete and regarding  what exam your asking.

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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